Solve for $x$ and $y$ by deriving an expression for $x$ from the second equation, and substituting it back into the first equation. $\begin{align*}-6x-3y &= 3 \\ 7x+4y &= -6\end{align*}$
Answer: Begin by moving the $y$ -term in the second equation to the right side of the equation. $7x = -4y-6$ Divide both sides by $7$ to isolate $x$ $x = {-\dfrac{4}{7}y - \dfrac{6}{7}}$ Substitute this expression for $x$ in the first equation. $-6({-\dfrac{4}{7}y - \dfrac{6}{7}}) - 3y = 3$ $\dfrac{24}{7}y + \dfrac{36}{7} - 3y = 3$ Simplify by combining terms, then solve for $y$ $\dfrac{3}{7}y + \dfrac{36}{7} = 3$ $\dfrac{3}{7}y = -\dfrac{15}{7}$ $y = -5$ Substitute $-5$ for $y$ in the top equation. $-6x-3( -5) = 3$ $-6x+15 = 3$ $-6x = -12$ $x = 2$ The solution is $\enspace x = 2, \enspace y = -5$.